Answer
$20^{\circ}$.
Work Step by Step
The question says that a projectile is launched upward at $70^{\circ}$, and asks for the other angle of launch that will produce the same range, for the same launch speed.
The answer in the back of the book says $15^{\circ}$ and refers you to Figure 10.11. Unfortunately, Figure 10.11 shows that $15^{\circ}$ produces the same range as $75^{\circ}$, not as $70^{\circ}$.
Since the figure in the book doesn't help answer the question for an initial angle of $70^{\circ}$, we turn to other resources. Equations in other texts or found via your favorite search engine (try "range equation") show that the range can be written as $R = \frac{(v_{initial})^{2}sin(2 \theta)}{g}$.
An initial angle of $70^{\circ}$ or $20^{\circ}$ gives the same range.
